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Hello there,

This is probably very easy but I can't find an argument.

Call a function $f: R^n \to R$ monotone increasing if $x_i \le y_i$ for each $1 \le i \le n$ implies $f(x) \le f(y)$.

I'd like to show that such a function is measurable; I'd be very surprised if this is not the case.

If $n=1$ it's ok, for the set $f^{-1}(-\infty,c)$ is either empty or an interval. I thought about studying each section of $f$, that is, fix e.g. $\hat{x}_1=(x_2,\ldots,x_n)$, then $f_{\hat{x}_1}(x) := f(x,x_2,\ldots,x_n)$ is monotone, hence measurable. Similarly any section is measurable. But it seems this is not sufficient to conclude...


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I think this appears as an exercise in Geoffrey Grimmett's Probability on Graphs. – Nate Eldredge Jun 21 '13 at 0:31
up vote 6 down vote accepted

We use induction on $n$. For $n=1$ this is trivial. Otherwise, suppose $f : \mathbb{R}^{n+1} \to \mathbb{R}$ is monotone increasing, fix $a \in \mathbb{R}$, and define $g : \mathbb{R}^n \to \mathbb{R}$ by $g(x) = \inf\{t \in \mathbb{R} : f(x,t) \ge a\}$. Then $g$ is monotone decreasing. By the induction hypothesis, $g$ is Lebesgue measurable. So the epigraph $E = \{(x,t) : t > g(x)\}$ is Lebesgue measurable, and the graph $G = \{(x,t) : g(x) = t\}$ is Lebesgue null (Fubini's theorem). However, $E \triangle\ \{f \ge a\} \subset G$ so $\{f \ge a\}$ is Lebesgue measurable.

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Thanks very much. – lonewolf Jun 21 '13 at 14:41

This does not work for Borel measurability even in $\mathbb{R}^{2}$. To see this, let $A\subseteq\mathbb{R}$, and let $R_{A}\subseteq\mathbb{R}$ be the set $\{(x,y)|-x<y\}\cup\{(x,-x)|x\in A\}$.` Then $R_{A}$ is an upward-closed subset of $\mathbb{R}^{2}$. Let $f_{A}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the function where $f_{A}(z)=1$ whenever $z\in R_{A}$ and $f_{A}(z)=0$ otherwise. Then $f_{A}$ is monotone increasing, but $f_{A}$ is Borel measurable if and only if $A$ is a Borel subset of $\mathbb{R}$. I suspect that every monotone map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is Lebesgue measurable in the $n$-dimensional Lebesgue measure, but I do not yet have a proof (I will get back to you if I find a proof). And Nate Eldredge beat me to the proof for the Lebesgue measurable case.

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Thanks, your example is interesting too and complements the story. I tried to accept both answers but it doesn't work. Thanks anyway :) Small typo: you meant $z \in R_A$. – lonewolf Jun 21 '13 at 10:35

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